Given an array nums of integers, return how many of them contain an even number of digits.
Example 1:
Input: nums = [12,345,2,6,7896] Output: 2 Explanation: 12 contains 2 digits (even number of digits). 345 contains 3 digits (odd number of digits). 2 contains 1 digit (odd number of digits). 6 contains 1 digit (odd number of digits). 7896 contains 4 digits (even number of digits). Therefore only 12 and 7896 contain an even number of digits.
Example 2:
Input: nums = [555,901,482,1771] Output: 1 Explanation: Only 1771 contains an even number of digits.
Constraints:
1 <= nums.length <= 5001 <= nums[i] <= 105
class Solution:
def findNumbers(self, nums: List[int]) -> int:
even_count = 0
for num in nums:
digit_count = int(math.floor(math.log10(num))) + 1
if digit_count%2 == 0:
even_count+=1
return even_count